From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:
A. 2 astronomical units.
B. 4 astronomical units.
C. 8 astronomical units.
D. 16 astronomical units.
E. It depends on the asteroid's mass.

Respuesta :

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

[tex]T^2\propto a^3\\\\T^2=ka^3[/tex]

It is mentioned that, an asteroid with an orbital period of 8 years. So,

[tex](8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU[/tex]

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.